Answer
$\det A=-5112$
Work Step by Step
Given $$\begin{array}{c}{A=\left[\begin{array}{ccc}
2 & {-1} & 3&4 \\
7 & {1} & 2&3 \\
-2&4 &8& 6\\
6&-6 &18 &-24\end{array}\right] }\end{array} $$
So, we get
$$A=\begin{vmatrix}
2 & {-1} & 3&4 \\
7 & {1} & 2&3 \\
-2&4 &8& 6\\
6&-6 &18 &-24
\end{vmatrix} \xrightarrow{R_2-\frac{7}{2}R_1,R_3+R_{1},R_4-3R_1} \begin{vmatrix}
2 & {-1} & 3&4 \\
0 & {\frac{9}{2}} & -5&-11 \\
0&3 &10& 10\\
0&-3&12 &-36
\end{vmatrix} \xrightarrow{R_{3}-\frac{2}{3}R_{2},R_4-2R_2,-\frac{33}{20}R_4-\frac{33}{20}R_3 }\begin{vmatrix}
2 & {-1} & 2&4 \\
0 & {\frac{9}{2}} & -5&-11 \\
0&0 &\frac{40}{3}&\frac{52}{3}\\
0&0 &0 &-\frac{213}{5}
\end{vmatrix} $$
\begin{array}{l}{\text { Determinant of triangular matrix is the product of its diagonal elements, so: }} \\ {\text {det }A=2 \cdot \frac{9}{2}\cdot \frac{40}{3}\cdot (-\frac{213}{5})=-5112}\end{array}
